Izmantojot ievades virkni un modeli, pārbaudiet, vai rakstzīmes ievades virknē atbilst tādai pašai secībai, ko nosaka shēmā esošās rakstzīmes. Pieņemsim, ka shēmā nebūs rakstzīmju dublikātu.
Ir publicēts vēl viens tās pašas problēmas risinājums šeit .
Piemēri:
Input: string = 'engineers rock' pattern = 'er'; Output: true All 'e' in the input string are before all 'r'. Input: string = 'engineers rock' pattern = 'egr'; Output: false There are two 'e' after 'g' in the input string. Input: string = 'engineers rock' pattern = 'gsr'; Output: false There are one 'r' before 's' in the input string.
Šeit doma ir samazināt doto virkni līdz dotajam modelim. Rakstzīmēm, kas norādītas shēmā, mēs saglabājam tikai atbilstošās rakstzīmes virknē. Jaunajā virknē mēs izdzēšam nepārtrauktas atkārtotas rakstzīmes. Pēc tam modificētajai virknei jābūt vienādai ar norādīto paraugu. Visbeidzot, mēs salīdzinām modificēto virkni ar norādīto modeli un attiecīgi atgriežam true vai false.
Ilustrācija:
tojson java
str = 'bfbaeadeacc' pat[] = 'bac' 1) Remove extra characters from str (characters that are not present in pat[] str = 'bbaaacc' [f e and d are removed] 3) Removed consecutive repeating occurrences of characters str = 'bac' 4) Since str is same as pat[] we return true
Zemāk ir aprakstīta iepriekš minēto darbību īstenošana.
// C++ code for the above approach #include
// Java program to check if characters of a string follow // pattern defined by given pattern. import java.util.*; public class OrderOfCharactersForPattern { public static boolean followsPattern(String str String pattern) { // Insert all characters of pattern in a hash set Set<Character> patternSet = neHashSet<>(); for (int i=0; i<pattern.length(); i++) patternSet.add(pattern.charAt(i)); // Build modified string (string with characters only from // pattern are taken) StringBuilder modifiedString = new StringBuilder(str); for (int i=str.length()-1; i>=0; i--) if (!patternSet.contains(modifiedString.charAt(i))) modifiedString.deleteCharAt(i); // Remove more than one consecutive occurrences of pattern // characters from modified string. for (int i=modifiedString.length()-1; i>0; i--) if (modifiedString.charAt(i) == modifiedString.charAt(i-1)) modifiedString.deleteCharAt(i); // After above modifications the length of modified string // must be same as pattern length if (pattern.length() != modifiedString.length()) return false; // And pattern characters must also be same as modified string // characters for (int i=0; i<pattern.length(); i++) if (pattern.charAt(i) != modifiedString.charAt(i)) return false; return true; } // Driver program int main() { String str = 'engineers rock'; String pattern = 'er'; System.out.println('Expected: true Actual: ' + followsPattern(str pattern)); str = 'engineers rock'; pattern = 'egr'; System.out.println('Expected: false Actual: ' + followsPattern(str pattern)); str = 'engineers rock'; pattern = 'gsr'; System.out.println('Expected: false Actual: ' + followsPattern(str pattern)); str = 'engineers rock'; pattern = 'eger'; System.out.println('Expected: true Actual: ' + followsPattern(str pattern)); return 0; } }
# Python3 program to check if characters of # a string follow pattern defined by given pattern. def followsPattern(string pattern): # Insert all characters of pattern in a hash set patternSet = set() for i in range(len(pattern)): patternSet.add(pattern[i]) # Build modified string (string with characters # only from pattern are taken) modifiedString = string for i in range(len(string) - 1 -1 -1): if not modifiedString[i] in patternSet: modifiedString = modifiedString[:i] + modifiedString[i + 1:] # Remove more than one consecutive occurrences # of pattern characters from modified string. for i in range(len(modifiedString) - 1 0 -1): if modifiedString[i] == modifiedString[i - 1]: modifiedString = modifiedString[:i] + modifiedString[i + 1:] # After above modifications the length of # modified string must be same as pattern length if len(pattern) != len(modifiedString): return False # And pattern characters must also be same # as modified string characters for i in range(len(pattern)): if pattern[i] != modifiedString[i]: return False return True # Driver Code if __name__ == '__main__': string = 'engineers rock' pattern = 'er' print('Expected: true Actual:' followsPattern(string pattern)) string = 'engineers rock' pattern = 'egr' print('Expected: false Actual:' followsPattern(string pattern)) string = 'engineers rock' pattern = 'gsr' print('Expected: false Actual:' followsPattern(string pattern)) string = 'engineers rock' pattern = 'eger' print('Expected: true Actual:' followsPattern(string pattern)) # This code is contributed by # sanjeev2552
// C# program to check if characters of a string follow // pattern defined by given pattern. using System; using System.Collections.Generic; using System.Text; class GFG { public static bool followsPattern(String str String pattern) { // Insert all characters of pattern in a hash set HashSet<char> patternSet = new HashSet<char>(); for (int i = 0; i < pattern.Length; i++) patternSet.Add(pattern[i]); // Build modified string (string with characters // only from pattern are taken) StringBuilder modifiedString = new StringBuilder(str); for (int i = str.Length - 1; i >= 0; i--) if (!patternSet.Contains(modifiedString[i])) modifiedString.Remove(i 1); // Remove more than one consecutive occurrences of pattern // characters from modified string. for (int i = modifiedString.Length - 1; i > 0; i--) if (modifiedString[i] == modifiedString[i - 1]) modifiedString.Remove(i 1); // After above modifications the length of modified string // must be same as pattern length if (pattern.Length != modifiedString.Length) return false; // And pattern characters must also be same // as modified string characters for (int i = 0; i < pattern.Length; i++) if (pattern[i] != modifiedString[i]) return false; return true; } // Driver program public static void Main(String[] args) { String str = 'engineers rock'; String pattern = 'er'; Console.WriteLine('Expected: true Actual: ' + followsPattern(str pattern)); str = 'engineers rock'; pattern = 'egr'; Console.WriteLine('Expected: false Actual: ' + followsPattern(str pattern)); str = 'engineers rock'; pattern = 'gsr'; Console.WriteLine('Expected: false Actual: ' + followsPattern(str pattern)); str = 'engineers rock'; pattern = 'eger'; Console.WriteLine('Expected: true Actual: ' + followsPattern(str pattern)); } } // This code is contributed by 29AjayKumar
<script> // Javascript program to check if characters of a string follow // pattern defined by given pattern. function followsPattern(str pattern) { // Insert all characters of pattern in a hash set let patternSet = new Set(); for (let i=0; i<pattern.length; i++) patternSet.add(pattern[i]); // Build modified string (string with characters only from // pattern are taken) let modifiedString = (str).split(''); for (let i=str.length-1; i>=0; i--) if (!patternSet.has(modifiedString[i])) modifiedString.splice(i1); // Remove more than one consecutive occurrences of pattern // characters from modified string. for (let i=modifiedString.length-1; i>0; i--) if (modifiedString[i] == modifiedString[i-1]) modifiedString.splice(i1); // After above modifications the length of modified string // must be same as pattern length if (pattern.length != modifiedString.length) return false; // And pattern characters must also be same as modified string // characters for (let i=0; i<pattern.length; i++) if (pattern[i] != modifiedString[i]) return false; return true; } // Driver program let str = 'engineers rock'; let pattern = 'er'; document.write('Expected: true Actual: ' + followsPattern(str pattern)+'
'); str = 'engineers rock'; pattern = 'egr'; document.write('Expected: false Actual: ' + followsPattern(str pattern)+'
'); str = 'engineers rock'; pattern = 'gsr'; document.write('Expected: false Actual: ' + followsPattern(str pattern)+'
'); str = 'engineers rock'; pattern = 'eger'; document.write('Expected: true Actual: ' + followsPattern(str pattern)+'
'); // This code is contributed by rag2127 </script>
Izvade:
Expected: true Actual: true Expected: false Actual: false Expected: false Actual: false Expected: true Actual: true
Laika sarežģītība: Iepriekš minēto implementāciju laika sarežģītība faktiski ir O(mn + n^2), jo mēs izmantojam deleteCharAt(), lai noņemtu rakstzīmes. Mēs varam optimizēt iepriekš minēto risinājumu, lai tas darbotos lineārā laikā. Tā vietā, lai izmantotu deleteCharAr (), mēs varam izveidot tukšu virkni un pievienot tai tikai nepieciešamās rakstzīmes.
StringBuilder tiek izmantots, lai darbotos ar ievades virkni. Tas ir tāpēc, ka StringBuilder ir mainīgs, savukārt String ir nemainīgs objekts. Lai izveidotu jaunu virkni, ir nepieciešams O(n) atstarpes, tāpēc papildu atstarpe ir O(n).
Mēs esam apsprieduši vēl divas pieejas šīs problēmas risināšanai.
Pārbaudiet, vai virkne atbilst rakstzīmju secībai, ko nosaka raksts, vai ne | 1. komplekts
Pārbaudiet, vai virkne atbilst rakstzīmju secībai, ko nosaka raksts, vai ne | 3. komplekts
kāpēc virkne ir nemainīga java